What is the correlation between $X_N$ and $Y_N$ as $N \to \infty$? are they asymptotically independent?

A quick simulation shows that this correlation drops as you increase $N$ but the decay is rather slow - so it's not clear if it goes to zero and if so, how rapidly. A possibly related result is that the max and sum of $N$ independent Gaussians are known to be asymptotically independent as $N \to \infty$ (see for example Ho, H. C. and Hsing, T. 1996).

1 Answer
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The main contribution to the correlation between $X_N$ and $Y_N$ is the event that the same $i$ maximizes $x_i$ and $y_i$. If $\rho$ is fixed, this event is asymptotically unlikely. (Given the value of $X_N=x_i$ we have that $E y_i=\rho X_N$, which is not large enough to make $y_i$ maximal.) For essentially the same reason they are asymptotically independent.

One way to make this precise is to first replace the number of samples by a Poisson with mean N. With high probability the resulting maxima are $X'_N=X_N$ and $Y'_N=Y_N$. However, this is now the rightmost and topmost points of a (non-homogeneous) Poisson process. These are w.h.p. located in different regions of the plane, so are almost independent.

Thanx! yes, I can prove that the prob. of the same index giving the maximum for $X_N$ and $Y_N$ goes to zero, and understand the mapping to Poisson process. Just to get a complete proof, is there a good reference for the fact that the two extreme points are independent? (perhaps this is standard/trivial - pardon my ignorance of point processes)
–
Or ZukApr 12 '11 at 22:36

You can identify two disjoint subsets of the plane so that w.h.p. the point maximizing x_i is in one and the point maximizing y_i is in the other. The restrictions of the Poisson process to disjoint sets are independent by definition.
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OmerApr 13 '11 at 15:26